Problem 168 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(S_{n}\) represents the sum of \(n\) terms of a G.P. whose first term and common ratio are \(a\) and \(r\) respectively, then prove that i. \(\quad S_{1}+S_{2}+S_{3}+\ldots+S_{n}=\frac{n a}{1-r}-\frac{\operatorname{ar}\left(1-r^{n}\right)}{(1-r)^{2}}\); ii. \(S_{1}+S_{3}+S_{5}+\ldots+S_{2 n-1}=\frac{a n}{1-r}-\frac{a r\left(1-r^{2 n}\right)}{(1-r)^{2}(1+r)}\).

Step-by-Step Solution

Answer

i. \(S_{1}+S_{2}+S_{3}+\ldots+S_{n}=\frac{n a}{1-r} - \frac{a r \left(1-r^{n}\right)}{(1-r)^2}\) \n\n ii. \(S_{1}+S_{3}+S_{5}+\ldots+S_{2n-1} = \frac{a n}{1-r} - \frac{a r\left(1-r^{2 n}\right)}{(1-r)^2 (1+r)}\)

Step 1: Start with first sum expression

Start by writing down the first sum equation and plugging in the formula for the sum of a geometric series. So we have \(S_{1}+S_{2}+S_{3}+\ldots+S_{n}\) = \(S_{1}+\frac{a(1-r^2)}{1-r}+S_{3}+\frac{a(1-r^3)}{1-r}+...+\frac{a(1-r^n)}{1-r}\).

Step 2: Simplifiying the equation

Simplifying this equation gives us: \[S_{1}+S_{2}+S_{3}+\ldots+S_{n} =\frac{na}{1-r} -\frac{a\cdot r (1+r+r^2+... +r^{n-1})}{(1-r)}\]

Step 3: Simplify further using the formula for the sum of a G.P.

The term inside brackets \((1+r+r^2+... +r^{n-1})\) is a geometric series with the first term as 1, a common ratio as \(r\) and number of terms as \(n\). \n\nSo applying the formula for the sum of \(n\) terms in a G.P. gives us \((1-r^n)/(1-r)\). Replacing it in the equation we obtain: \[S_{1}+S_{2}+S_{3}+\ldots+S_{n} = \frac{n a}{1-r} - \frac{a r \left(1-r^{n}\right)}{(1-r)^2}\]

Step 4: Start with the second sum expression

Like in step 1, to solve the second part of the problem write down the second sum equation and substitute the formula for the sum \(S_{2n-1}\) at each indexed position: \[S_{1}+S_{3}+...\ldots+S_{2n-1} = S_{1}+\frac{a(1-r^3)}{1-r}+\frac{a(1-r^5)}{1-r}+...+\frac{a(1-r^{2n-1})}{1-r}\]

Step 5: Simplifying second sum equation

Simplifying this equation we obtain: \[S_{1}+S_{3}+...+\ldots+S_{2n-1} = \frac{an}{1-r} -\frac{a\cdot r (1+r^2+r^4+...+r^{2n-2})}{(1-r)}\]

Step 6: Simplify further using formula for sum of a G.P.

The term inside brackets \((1+r^2+r^4+... +r^{2n-2})\) is a geometric series with first term as 1, common ratio as \(r^2\) and number of terms as \(n\). \n\nApplying the formula for sum of \(n\) terms in a G.P. we then get \((1-r^{2n})/(1-r^2)\). Replacing it in the equation we obtain: \[S_{1}+S_{3}+S_{5}+\ldots+S_{2n-1} = \frac{a n}{1-r} - \frac{a r\left(1-r^{2 n}\right)}{(1-r)^2 (1+r)}\] which completes the second part of the proof.

Key Concepts

Sum of a GP

In the realm of sequences, the geometric progression (GP) holds a pivotal spot, especially when it comes to understanding sums of sequences. A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio, denoted as \( r \). The formula for the sum of the first \( n \) terms of a GP where the first term is \( a \) is given by:

• If \( r eq 1 \), then \( S_n = a \frac{1-r^n}{1-r} \).
• If \( r = 1 \), then the series actually becomes arithmetic, and \( S_n = n \cdot a \).

The provided exercise challenges us to use this sum formula in more complex ways to prove given expressions involving sums of specific terms of a GP. The exercise aims to deepen the understanding of manipulating series not just individually but inter-connectedly by levels (\( S_1 + S_2 + S_3 + \ldots + S_n \)) and by selecting every alternate term (\( S_1 + S_3 + S_5 + \ldots + S_{2n-1} \)).
By carefully applying the sum formula, we see how these complex expressions become manageable, leading to tidy proofs of the summation identities in the original problem statement.

Series and Sequences

Series and sequences are fundamental concepts in mathematics. Understanding their properties is crucial for both theoretical and practical applications. A sequence is simply an ordered list of numbers, while a series is the sum of the terms of a sequence.
In the context of this exercise, a geometric series is specifically being dealt with, which is significant for its multiplicative structure. With a sequence defined by its terms and a formula to find the sum, it forms the basis for solving more complicated problems within mathematics and related fields.

• Each GP sequence is defined by the first term \( a \) and the common ratio \( r \).
• The challenge often lies not only in finding the sum but understanding how terms relate and interact within their positioning, such as in partial sums \( S_1, S_2, \ldots \).

In our problem, sequences are specifically synthetized into partial sums and alternate sums, teasing apart the interactions between terms multiplied and distributed over a series.
Ultimately, mastering series and sequences involves spotting these patterns and strategically employing known formulas to simplify or transform them.

Proof Techniques

Proof techniques are essential tools in mathematics for establishing the truth of given propositions. In the context of this exercise, we employ algebraic manipulation and known formulas to demonstrate the given identities involving sums of terms in a GP.
The step-by-step solution showcases one particular style of proof, often employed in algebraic scenarios, namely:

• Start with the known formula: Begin by applying the sum formula for a GP \( S_n = a \frac{1-r^n}{1-r} \).
• Simplify expressions: Use algebraic manipulation to simplify or relate parts of the expression. Recognize given series patterns like \( r^k \) in terms or sums \( (1+r+r^2+...+r^{n-1}) \) which are also smaller GPs.
• Substitution and rearrangement: Substitute back into the larger expression to arrive at the result.

These proof techniques help not only to prove the results but also to provide deeper insight into why the formulas work and how such expression patterns arise.
The visual breakdown and logical flow of these processes form bridges between intuitive understanding and rigorous mathematical reasoning.